# ICSE Solutions for Selina Concise Chapter 5 Quadratic Equation Class 10 Maths

### Exercise 5(A)

1. Find which of the following equations are quadratic:

(i) (3x – 1)2 = 5(x + 8)

(ii) 5x2 – 8x = -3(7 – 2x)

(iii) (x -4) (3x + 1) = (3x – 1) (x + 2)

(iv) x2 + 5x – 5 = (x – 3)2

(v) 7x3 – 2x2 + 10 = (2x – 5)2

(vi) (x – 1)2 + (x + 2)2 + 3(x + 1) = 0

Solution

(i) (3x – 1)2 = 5(x + 8)

⇒ (9x2 – 6x + 1) = 5x + 40

⇒ 9x2 – 11x – 39 = 0; which is of the general form ax2 + bx + c = 0.

Thus, the given equation is a quadratic equation.

(ii) 5x2 – 8x = -3(7 – 2x)

⇒ 5x2 – 8x = 6x – 21

⇒ 5x2 – 14x + 21 = 0; which is of the general form ax2 + bx + c = 0.

Thus, the given equation is a quadratic equation.

(iii) (x – 4) (3x + 1) = (3x – 1) (x +2)

⇒ 3x2 + x – 12x – 4 = 3x2 + 6x – x – 2

⇒ 16x + 2 = 0; which is not of the general form ax2 + bx + c = 0. And it’s a linear equation.

Thus, the given equation is not a quadratic equation.

(iv) x2 + 5x – 5 = (x – 3)2

⇒ x2 + 5x – 5 = x2 – 6x + 9

⇒ 11x – 14 = 0; which is not of the general form ax2 + bx + c = 0. And it’s a linear equation.

Thus, the given equation is not a quadratic equation.

(v) 7x3 – 2x2 + 10 = (2x – 5)2

⇒ 7x3 – 2x2 + 10 = 4x2 – 20x + 25

⇒ 7x3 – 6x2 + 20x – 15 = 0; which is not of the general form ax2 + bx + c = 0. And it’s a cubic equation.

Thus, the given equation is not a quadratic equation.

(vi) (x – 1)2 + (x + 2)2 + 3(x +1) = 0

⇒ x2 – 2x + 1 + x2 + 4x + 4 + 3x + 3 = 0

⇒ 2x2 + 5x + 8 = 0; which is of the general form ax2 + bx + c = 0.

Thus, the given equation is a quadratic equation.

2. (i) Is x = 5 a solution of the quadratic equation x2 – 2x – 15 = 0?

Solution

Given quadratic equation, x2 – 2x – 15 = 0

We know that, for x = 5 to be a solution of the given quadratic equation it should satisfy the equation.

Now, on substituting x = 5 in the given equation, we have

L.H.S = (5)2 – 2(5) – 15

= 25 – 10 – 15

= 0

= R.H.S

∴ x = 5 is a solution of the given quadratic equation x2 – 2x – 15 = 0

(ii) Is x = -3 a solution of the quadratic equation 2x2 – 7x + 9 = 0?

Solution

Given quadratic equation, 2x2 – 7x + 9 = 0

We know that, for x = -3 to be solution of the given quadratic equation it should satisfy the equation.

Now, on substituting x = 5 in the given equation, we have

L.H.S = 2(-3)2 – 7(-3) + 9

= 18 + 21 + 9

= 48

≠ R.H.S

∴ x = -3 is not a solution of the given quadratic equation 2x2 – 7x + 9 = 0.

### Exercise 5(B)

1. Without solving, comment upon the nature of roots of each of the following equations:

(i) 7x2 – 9x +2 = 0

(ii) 6x2 – 13x +4 = 0

(iii) 25x2 – 10x + 1= 0

(iv) x2 + 2√3x – 9 = 0

(v) x2 – ax – b2 = 0

(vi) 2x2 + 8x + 9 = 0

Solution

(i) Given quadratic equation, 7x2 – 9x + 2 = 0

Here, a = 7, b = -9 and c = 2

So, the Discriminant (D) = b2 – 4ac

D = (-9)2 – 4(7)(2)

= 81 – 56

= 25

As D > 0, the roots of the equation is real and unequal.

(ii) Given quadratic equation, 6x2 – 13x + 4 = 0

Here, a = 6, b = -13 and c = 4

So, the Discriminant (D) = b2 – 4ac

D = (-13)2 – 4(6)(4)

= 169 – 96

= 73

As D > 0, the roots of the equation is real and unequal.

(iii) Given quadratic equation, 25x2 – 10x + 1 = 0

Here, a = 25, b = -10 and c = 1

So, the Discriminant (D) = b2 – 4ac

D = (-10)2 – 4(25)(1)

= 100 – 100

= 0

As D = 0, the roots of the equation is real and equal.

(iv) Given quadratic equation, x2 + 2√3x – 9 = 0

Here, a = 1, b = 2√3 and c = -9

So, the Discriminant (D) = b2 – 4ac

D = (2√3)2 – 4(1)(-9)

= 12 + 36

= 48

As D > 0, the roots of the equation is real and unequal.

(v) Given quadratic equation, x2 – ax – b2 = 0

Here, a = 1, b = -a and c = -b2

So, the Discriminant (D) = b2 – 4ac

D = (a)2 – 4(1)(-b2)

= a2 + 4b2

a2 + 4b2 is always positive value.

Thus D > 0, and the roots of the equation is real and unequal

(vi) Given quadratic equation, 2x2 + 8x + 9 = 0

Here, a = 2, b = 8 and c = 9

So, the Discriminant (D) = b2 – 4ac

D = (8)2 – 4(2)(9)

= 64 – 72

= -8

As D < 0, the equation has no roots.

2. Find the value of ‘p’, if the following quadratic equations has equal roots:

(i) 4x2 – (p – 2)x + 1 = 0

(ii) x2 + (p – 3)x + p = 0

Solution

(i) 4x2 – (p – 2)x + 1 = 0

Here, a = 4, b = -(p – 2), c = 1

Given that the roots are equal,

So, Discriminant = 0 ⇒ b2– 4ac = 0

D = (-(p – 2))– 4(4)(1) = 0

⇒ p2 + 4 – 4p – 16 = 0

⇒ p2 – 4p – 12 = 0

⇒ p2 – 6p + 2p – 12 = 0

⇒ p(p – 6) + 2(p – 6) = 0

⇒ (p + 2)(p – 6) = 0

⇒ p + 2 = 0 or p – 6 = 0

Hence, p = -2 or p = 6

(ii) x2 + (p – 3)x + p = 0

Here, a = 1, b = (p – 3), c = p

Given that the roots are equal,

So, Discriminant = 0 ⇒ b2– 4ac = 0

D = (p – 3)– 4(1)(p) = 0

⇒ p2 + 9 – 6p – 4p = 0

⇒ p2– 10p + 9 = 0

⇒ p2-9p – p + 9 = 0

⇒ p(p – 9) – 1(p – 9) = 0

⇒ (p -9)(p – 1) = 0

⇒ p – 9 = 0 or p – 1 = 0

Hence, p = 9 or p = 1

### Exercise 5(C)

Solve equations, number 1 to 20, given below, using factorization method:

1. x2 – 10x – 24 = 0

Solution

Given equation, x2 – 10x – 24 = 0

⇒ x2 – 12x + 2x – 24 = 0

⇒ x(x – 12) + 2(x – 12) = 0

⇒ (x + 2)(x – 12) = 0

So, x + 2 = 0 or x – 12 = 0

Hence, x = -2 or x = 12

2. x2 – 16 = 0

Solution

Given equation, x2 – 16 = 0

⇒ x2 + 4x – 4x + 16 = 0

⇒ x(x + 4) -4(x + 4) = 0

⇒ (x – 4) (x + 4) = 0

So, (x – 4) = 0 or (x + 4) = 0

Hence, x = 4 or x = -4

3. 2x2 – ½ x = 0

Solution

Given equation, 2x2 – ½ x = 0

⇒ 4x2 – x = 0

⇒ x(4x – 1) = 0

So, either x = 0 or 4x – 1 = 0

Hence, x = 0 or x = ¼

4. x(x – 5) = 24

Solution

Given equation, x(x – 5) = 24

⇒ x2 – 5x = 24

⇒ x2 – 5x – 24 = 0

⇒ x2 – 8x + 3x – 24 = 0

⇒ x(x – 8) + 3(x – 8) = 0

⇒ (x + 3)(x – 8) = 0

So, x + 3 = 0 or x – 8 = 0

Hence, x = -3 or x = 8

5. 9/2 x = 5 + x2

Solution

Given equation, 9/2 x = 5 + x2

On multiplying by 2 both sides, we have

9x = 2(5 + x2)

⇒ 9x = 10 + 2x2

⇒ 2x2 – 9x + 10 = 0

⇒ 2x2 – 4x – 5x + 10 = 0

⇒ 2x(x – 2) – 5(x – 2) = 0

⇒ (2x – 5)(x -2) = 0

So, 2x – 5 = 0 or x – 2 = 0

Hence, x = 5/2 or x = 2

6. 6/x = 1 + x

Solution

Given equation, 6/x = 1 + x

On multiplying by x both sides, we have

6 = x(1 + x)

⇒ 6 = x + x2

⇒ x2 + x – 6 = 0

⇒ x2 + 3x – 2x – 6 = 0

⇒ x(x + 3) – 2(x + 3) = 0

⇒ (x – 2) (x + 3) = 0

So, x – 2 = 0 or x + 3 = 0

Hence, x = 2 or x = -3

7. x = (3x + 1)/ 4x

Solution

Given equation, x = (3x + 1)/ 4x

On multiplying by 4x both sides, we have

4x(x) = 3x + 1

⇒ 4x2 = 3x + 1

⇒ 4x2 – 3x – 1 = 0

⇒ 4x2 – 4x + x – 1 = 0

⇒ 4x(x – 1) + 1(x – 1) = 0

⇒ (4x + 1) (x – 1) = 0

So, 4x + 1 = 0 or x – 1 = 0

Hence, x = -1/4 or x = 1

8. x + 1/x = 2.5

Solution

Given equation, x + 1/x = 2.5

x + 1/x = 5/2

Taking LCM on L.H.S, we have

(x2 + 1)/ x = 5/2

⇒ 2(x2 + 1) = 5x

⇒ 2x2 + 2 = 5x

⇒ 2x2 – 5x + 2 = 0

⇒ 2x2 – 4x – x + 2 = 0

⇒ 2x(x – 2) -1(x – 2) = 0

⇒ (2x – 1)(x – 2) = 0

So, 2x – 1 = 0 or x – 2 = 0

Hence, x = ½ or x = 2

9. (2x – 3)2 = 49

Solution

Given equation, (2x – 3)2 = 49

Expanding the L.H.S, we have

4x2 – 12x + 9 = 49

⇒ 4x2 – 12x – 40 = 0

Dividing by 4 on both side

x2 – 3x – 10 = 0

⇒ x2 – 5x + 2x – 10 = 0

⇒ x(x – 5) + 2(x – 5) = 0

⇒ (x + 2) (x – 5) = 0

So, x + 2 = 0 or x – 5 = 0

Hence, x = -2 or 5

10. 2(x2 – 6) = 3(x – 4)

Solution

Given equation, 2(x2 – 6) = 3(x – 4)

2x2 – 12 = 3x – 12

⇒ 2x2 = 3x

⇒ x(2x – 3) = 0

So, x = 0 or (2x – 3) = 0

Hence, x = 0 or x = 3/2

11. (x + 1) (2x + 8) = (x + 7) (x + 3)

Solution

Given equation, (x + 1) (2x + 8) = (x + 7) (x + 3)

2x2 + 2x + 8x + 8 = x2 + 7x + 3x + 21

⇒ 2x2 + 10x + 8 = x2 + 10x + 21

⇒ x= 21 – 8

⇒ x– 13 = 0

⇒ (x – √13) (x + √13) = 0

So, x – √13 = 0 or x + √13 = 0

Hence, x = – √13 or x = √13

12. x2 – (a + b)x + ab = 0

Solution

Given equation, x2 – (a + b)x + ab = 0

x2 – ax – bx + ab = 0

⇒ x(x – a) – b(x – a) = 0

⇒ (x – b) (x – a) = 0

So, x – b = 0 or x – a = 0

Hence, x = b or x = a

13. (x + 3)2 – 4(x + 3) – 5 = 0

Solution

Given equation, (x + 3)2 – 4(x + 3) – 5 = 0

⇒ (x2 + 9 + 6x) – 4x – 12 – 5 = 0

⇒ x2 + 2x – 8 = 0

⇒ x2 + 4x – 2x – 8 = 0

⇒ x(x + 4) – 2(x – 4) = 0

⇒ (x – 2)(x + 4) = 0

So, x – 2 = 0 or x + 4 = 0

Hence, x = 2 or x = -4

14. 4(2x – 3)2 – (2x – 3) – 14 = 0

Solution

Given equation, 4(2x – 3)2 – (2x – 3) – 14 = 0

Let substitute 2x – 3 = y

Then the equation becomes,

4y2 – y – 14 = 0

⇒ 4y2 – 8y + 7y – 14 = 0

⇒ 4y(y – 2) + 7(y – 2) = 0

⇒ (4y + 7)(y – 2) = 0

So, 4y + 7 = 0 or y – 2 = 0

Hence, y = -7/4 or y = 2

But we have taken y = 2x – 3

Thus,

2x – 3 = -7/4 or 2x – 3 = 2

⇒ 2x = 5/ 4 or 2x = 5

⇒ x = 5/8 or x = 5/2

15. 3x – 2/ 2x- 3 = 3x – 8/ x + 4

Solution

Given equation, 3x – 2/ 2x- 3 = 3x – 8/ x + 4

On cross-multiplying we have,

(3x – 2)(x + 4) = (3x – 8)(2x – 3)

⇒ 3x2 – 2x + 12x – 8 = 6x2 – 16x – 9x + 24

⇒ 3x+ 10x – 8 = 6x2 – 25x + 24

⇒ 3x2 – 35x + 32 = 0

⇒ 3x2 – 3x – 32x + 32 = 0

⇒ 3x(x – 1) – 32(x – 1) = 0

⇒ (3x – 32)(x – 1) = 0

So, 3x – 32 = 0 or x – 1 = 0

Hence, x = 32/3 or x = 1

16. 2x2 – 9x + 10 = 0, when:

(i) x ∈ N

(ii) x ∈ Q

Solution

Given equation, 2x2 – 9x + 10 = 0

2x2 – 4x – 5x + 10 = 0

⇒ 2x(x – 2) – 5(x – 2) = 0

⇒ (2x – 5)(x – 2) = 0

So, 2x – 5 = 0 or x – 2 = 0

Hence, x = 5/2 or x = 2

(i) When x ∈ N

x = 2 is the solution.

(ii) When x ∈ Q

x = 2, 5/2 are the solutions

17. (x-3)/(x+3) + (x+3)/(x-3) = 2(1/2)

Solution

⇒ 2(2x2 + 18) = 5(x2 – 9)

⇒ 4x2 + 36 = 5x2 – 45

⇒ x2 – 81 = 0

⇒ (x – 9)(x + 9) = 0

So, x – 9 = 0 or x + 9 = 0

Hence, x = 9 or x = -9

### Exercise 5(D)

1. Solve, each of the following equations, using the formula:

(i) x2 – 6x = 27

Solution

Given equation, x2 – 6x = 27

x2 – 6x – 27 = 0

Here, a = 1 , b = -6 and c = -27

∴ x = 9 or -3

(ii) x2 – 10x + 21 = 0

Solution

Given equation, x2 – 10x + 21 = 0

Here, a = 1, b = -10 and c = 21

∴ x = 7 or x = 3

(iii) x2 + 6x – 10 = 0

Solution

Given equation, x2 + 6x – 10 = 0

Here, a = 1, b = 6 and c = -10

∴ x = -3 + √19 or x = -3 – √19

(iv) x2 + 2x – 6 = 0

Solution

Given equation, x2 + 2x – 6 = 0

Here, a = 1, b = 2 and c = -6

∴ x = -1 + √7 or x = -1 – √7

(v) 3x2 + 2x – 1 = 0

Solution

Given equation, 3x2 + 2x – 1 = 0

Here, a = 3, b = 2 and c = -1

∴ x = 1/3 or x = -1

(vi) 2x2 + 7x + 5 = 0

Solution

Given equation, 2x2 + 7x + 5 = 0

Here, a = 2, b = 7 and c = 5

∴ x = -1 or x = -5/2

(vii) 2/3 x = -1/6 x2 – 1/3

Solution

Given equation, 2/3 x = -1/6 x2 – 1/3

1/6 x2 + 2/3 x + 1/3 = 0

Multiplying by 6 on both sides

x2 + 4x + 2 = 0

Here, a = 1, b = 4 and c = 2

∴ x = -2 + √2 or x = -2 – √2

(viii) 1/15 x2 + 5/3 = 2/3 x

Solution

Given equation, 1/15 x2 + 5/3 = 2/3 x

1/15 x2 – 2/3 x + 5/3 = 0

Multiplying by 15 on both sides

x2 – 10x + 25 = 0

Here, a = 1, b = -10 and c = 25

∴ x = 5 (equal roots)

(ix) x2 – 6 = 2 √2 x

Solution

Given equation, x2 – 6 = 2 √2 x

x2 – 2√2 x – 6 = 0

Here, a = 1, b = -2√2 and c = -6

∴ x = 3√2 or x = -√2

(x) 4/x – 3 = 5/ (2x + 3)

Solution

Given equation, 4/x – 3 = 5/ (2x + 3)

(4 – 3x)/ x = 5/ (2x + 3)

On cross multiplying, we have

(4 – 3x)(2x + 3) = 5x

⇒ 8x – 6x2 + 12 – 9x = 5x

⇒ 6x2 + 6x – 12 = 0

Dividing by 6, we get

x+ x – 2 = 0

Here, a = 1, b = 1 and c = -2

∴ x = 1 or x = -2

(xi) 2x + 3/ x + 3 = x + 4/ x + 2

Solution

Given equation, 2x + 3/ x + 3 = x + 4/ x + 2

On cross-multiplying, we have

(2x + 3) (x + 2) = (x + 4) (x + 3)

⇒ 2x2 + 4x + 3x + 6 = x2 + 3x + 4x + 12

⇒ 2x2 + 7x + 6 = x+ 7x + 12

⇒ x2 + 0x – 6 = 0

Here, a = 1, b = 0 and c = -6

∴ x = √6 or x = -√6

(xii) √6x2 – 4x – 2√6 = 0

Solution

Given equation, √6x2 – 4x – 2√6 = 0

Here, a = √6, b = -4 and c = -2√6

∴ x = √6 or -√6/3

Solution

⇒ 25x2 – 175x + 300 = 12x2 – 57x + 60

⇒ 13x2 – 118x + 240 = 0

Here, a = 13, b = -118 and c = 240

∴ x = 6 or x = 40/13

(xiv) (x-1)/(x-2) + (x-3)/(x-4) = 3(1/3)

Solution

From the given equation,

⇒ 10x2 – 60x + 80 = 6x2 – 30x + 30

⇒ 4x2 – 30x + 50 = 0

⇒ 2x2 – 15x + 25 = 0

Here, a = 2, b = -15 and c = 25

∴ x = 5 or x = 5/2

2. Solve each of the following equations for x and give, in each case, your answer correct to one decimal place:

(i) x2 – 8x +5 = 0

(ii) 5x2 + 10x – 3 = 0

Solution

(i) x2 – 8x + 5 = 0

Here, a = 1, b = -8 and c = 5

x = 4 ± 3.3

Thus, x = 7.7 or x = 0.7

(ii) 5x2 + 10x – 3 = 0

Here, a = 5, b = 10 and c = -3

Thus, x = 0.3 or x = -2.3

3. Solve each of the following equations for x and give, in each case, your answer correct to 2 decimal places:

(i) 2x2 – 10x + 5 = 0

Solution

Given equation, 2x2 – 10x + 5 = 0

Here, a = 2, b = -10 and c = 5

∴ x = 4.44 or x = 0.56

(ii) 4x + 6/x + 13 = 0

Solution

Given equation, 4x + 6/x + 13 = 0

Multiplying by x both sides, we get

4x2 + 13x + 6 = 0

Here, a = 4, b = 13 and c = 6

∴ x = -0.56 or x = -2.70

(iii) 4x2 – 5x – 3 = 0

Solution

Given equation, 4x2 – 5x – 3 = 0

Here, a = 4, b = -5 and c = -3

∴ x = 1.70 or x = -0.44

(iv) x2 – 3x – 9 = 0

Solution

Given equation, x2 – 3x – 9 = 0

Here, a = 1, b = -3 and c = -9

∴ x = 4.85 or x = -1.85

(v) x– 5x – 10 = 0

Solution

Given equation, x– 5x – 10 = 0

Here, a = 1, b = -5 and c = -10

∴ x = 6.53 or x = -1.53

4. Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places:

(i) 3x2 – 12x – 1 = 0

(ii) x2 – 16 x +6 = 0

(iii) 2x2 + 11x + 4 = 0

Solution

(i) Given equation, 3x2 – 12x – 1 = 0

Here, a = 3, b = -12 and c = -1

∴ x = 4.082 or x = -0.082

(ii) Given equation, x2 – 16 x + 6 = 0

Here, a = 1, b = -16 and c = 6

∴ x = 15.616 or x = 0.384

(iii) Given equation, 2x2 + 11x + 4 = 0

Here, a = 2, b = 11 and c = 4

∴ x = -0.392 or x = -5.110

5. Solve:

(i) x4 – 2x2 – 3 = 0

Solution

Given equation, x4 – 2x2 – 3 = 0

⇒ x4 – 3x2 + x– 3 = 0

⇒ x2(x2 – 3) + 1(x– 3) = 0

⇒ (x2 + 1) (x2 – 3) = 0

So, x2 + 1 = 0 (which is not possible) or x2 – 3 = 0

Hence, x2 – 3 = 0

⇒ x = ± √3

(ii) x4 – 10x2 + 9 = 0

Solution

Given equation, x4 – 10x2 + 9 = 0

⇒ x4 – x2 – 9x2+ 9 = 0

⇒ x2(x– 1) – 9(x2 – 1) = 0

⇒ (x2 – 9)(x2 – 1) = 0

So, we have

x2 – 9 = 0 or x2 – 1 = 0

Hence, x = ± 3 or x = ± 1

### Exercise 5(E)

1. Solve each of the following equations:

2x/(x-3) + 1/(2x+3) + (3x+9)/(x-3)(2x+3) = 0; x ≠ 3, x ≠ -3/2

Solution

Given equation,

⇒ 4x2 + 6x + x – 3 + 3x + 9 = 0

⇒ 4x2 + 10x + 6 = 0

⇒ 4x2 + 4x + 6x + 6 = 0

⇒ 4x(x + 1) + 6(x + 1) = 0

⇒ (4x + 6) (x + 1) = 0

So, 4x + 6 = 0 or x + 1 = 0

x = -1 or x = -6/4 = -3/2 (rejected as this value is excluded in the domain)

∴ x = -1 is the only solution

2. (2x + 3)2 = 81

Solution

Given, (2x + 3)2 = 81

Taking square root on both sides we have,

2x + 3 = ± 9

⇒ 2x = ± 9 – 3

⇒ x = (± 9 – 3)/ 2

So, x = (9 – 3)/ 2 or (-9 – 3)/ 2

∴ x = 3 or x = -6

3. a2x2 – b2 = 0

Solution

Given equation, a2x2 – b2 = 0

⇒ (ax)2 – b2 = 0

⇒ (ax + b)(ax – b) = 0

So, ax + b = 0 or ax – b = 0

∴ x = -b/a or b/a

4. x2 – 11/4 x + 15/8 = 0

Solution

Given equation, x2 – 11/4 x + 15/8 = 0

Taking L.C.M we have,

(8x2 – 22x + 15)/ 8 = 0

⇒ 8x2 – 22x + 15 = 0

⇒ 8x2 – 12x – 10x + 15 = 0

⇒ 4x(2x – 3) – 5(2x – 3) = 0

⇒ (4x – 5)(2x – 3) = 0

So, 4x – 5 = 0 or 2x – 3 = 0

∴ x = 5/4 or x = 3/2

5. x + 4/x = -4; x ≠ 0

Solution

Given equation, x + 4/x = -4

⇒ (x+ 4)/ x = -4

⇒ x2 + 4 = -4x

⇒ x2 + 4x + 4 = 0

⇒ x2 + 2x + 2x + 4 = 0

⇒ x(x + 2) + 2(x + 2) = 0

⇒ (x + 2)(x + 2) = 0

⇒ (x + 2)2 = 0

Taking square–root we have,

x + 2 = 0

∴ x = -2

6. 2x4 – 5x2 + 3 = 0

Solution

Given equation, 2x4 – 5x2 + 3 = 0

Let’s take x2 = y

Then, the equation becomes

2y2 – 5y + 3 = 0

⇒ 2y2 – 2y – 3y + 3 = 0

⇒ 2y(y – 1) – 3(y – 1) = 0

⇒ (2y – 3) (y – 1) = 0

So, 2y – 3 = 0 or y – 1 = 0

y = 3/2 or y = 1

And, we have taken y = x2

Thus,

x2 = 3/2 or x2 = 1

⇒ x = ± √(3/2) or x = ±1

7. x4 – 2x2 – 3 = 0

Solution

Given equation, x4 – 2x2 – 3 = 0

Let’s take x2 = y

Then, the equation becomes

y2 – 2y – 3 = 0

⇒ y2 – 3y + y – 3 = 0

⇒ y(y – 3) + 1(y – 3) = 0

⇒ (y + 1)(y – 3) = 0

So, y + 1 = 0 or y – 3 = 0

⇒ y = -1 or y = 3

And, we have taken y = x2

Thus, x2 = – 1(impossible, no real solution)

⇒ x2 = 3

⇒ x = ± √3

8. 9(x2 + 1/x2)− 9(x+1/x) – 52 = 0

Solution

Let us take (x + 1/x) = y …. (1)

Now, squaring it on both sides

(x + 1/x)2 = y2

⇒ x2 + 1/x2 + 2 = y2

So, x2 + 1/x2 = y2 – 2 ... (2)

Using (1) and (2) in the given equation, we have

9(y2 – 2) – 9(y) – 52 = 0

⇒ 9y2 – 18 – 9y – 52 = 0

⇒ 9y2 – 9y – 70 = 0

⇒ 9y2 – 30y + 21y – 70 = 0

⇒ 3y(3y – 10) + 7(3y – 10) = 0

⇒ (3y + 7)(3y – 10) = 0

So, 3y + 7 = 0 or 3y – 10 = 0

⇒ y = -7/3 or y = 10/3

Now,

x + 1/x = -7/3 or x + 1/x = 10/3

⇒ (x2 + 1)/x = -7/3 or (x2 + 1)/x = 10/3

⇒ 3x2 – 10x + 3 = 0 or 3x2 + 7x + 3 = 0

⇒ 3x2 – 9x – x + 3 = 0 or

⇒ 3x(x – 3) – 1(x – 3) = 0

⇒ (3x – 1)(x – 3) = 0

So, x = 1/3 or 3

9. 2(x2 + 1/x2)− (x+1/x) = 11

Solution

Let us take (x + 1/x) = y …. (1)

Now, squaring it on both sides

(x + 1/x)2 = y2

⇒ x2 + 1/x2 + 2 = y2

So,

x2 + 1/x2 = y2 – 2 ….. (2)

Using (1) and (2) in the given equation, we have

2(y2 – 2) – (y) = 11

⇒ 2y2 – 4 – y = 11

⇒ 2y2 – y – 15 = 0

⇒ 2y2 – 6y + 5y – 15 = 0

⇒ 2y(y – 3) + 5(y – 3) = 0

⇒ (2y + 5) (y – 3) = 0

So, 2y + 5 = 0 or y – 3 = 0

⇒ y = -5/2 or y = 3

Now, x + 1/x = -5/2 or x + 1/x = 3

⇒ (x2 + 1)/x = -5/2 or (x2 + 1)/x = 3

⇒ 2(x2 + 1) = -5x or x2 + 1 = 3x

2x2 + 5x + 2 = 0 or x– 3x + 1 = 0

⇒ 2x2 + 4x + x + 2 = 0 or

⇒ 2x(x + 2) + 1(x + 2) = 0

⇒ (2x + 1)(x + 2) = 0

Hence, x = -1/2 or -2

10. (x2 + 1/x2)− 3(x−1/x) −2 = 0

Solution

Let us take (x – 1/x) = y …. (1)

Now, squaring it on both sides

(x – 1/x)2 = y2

⇒ x2 + 1/x2 – 2 = y2

So, x2 + 1/x2 = y2 + 2 ….. (2)

Using (1) and (2) in the given equation, we have

(y2 + 2) – 3(y) – 2 = 0

⇒ y2 -3y = 0

⇒ y(y – 3) = 0

So, y = 0 or y – 3 = 0

Now,

(x – 1/x) = 0 or (x – 1/x) = 3

⇒ x2 – 1 = 0 or x2 – 1 = 3x

⇒ x2 = 1 or x2 – 3x – 1 = 0

∴ x = ± 1

### Exercise 5(F)

1. Solve:

(i) (x + 5) (x – 5) = 24

Solution

Given equation, (x + 5) (x – 5) = 24

⇒ x2 – 25 = 24

⇒ x2 = 49

Thus, x = ± 7

(ii) 3x2 – 2√6x + 2 = 0

Solution

Given equation, 3x2 – 2√6x + 2 = 0

⇒ 3x2 – √6x – √6x + 2 = 0

⇒ √3x(√3x – √2) – √2(√3x – √2) = 0

⇒ (√3x – √2) (√3x – √2) = 0

So, √3x – √2 = 0 or √3x – √2 = 0

∴ x = √(2/3), √(2/3) (equal roots)

(iii) 3√2x2 – 5x – √2 = 0

Solution

Given equation, 3√2x2 – 5x – √2 = 0

⇒ 3√2x2 – 6x + x – √2 = 0

⇒ 3√2x(x – √2) + 1(x – √2) = 0

⇒ (3√2x + 1) (x – √2) = 0

So, 3√2x + 1 = 0 or x – √2 = 0

∴ x = -1/ 3√2 or x = √2

(iv) 2x – 3 = √(2x2 – 2x + 21)

Solution

Given equation, 2x – 3 = √(2x2 – 2x + 21)

On squaring on both sides, we have

(2x – 3)2 = 2x2 – 2x + 21

⇒ 4x2 + 9 – 12x = 2x2 – 2x + 21

⇒ 2x2 – 10x – 12 = 0

Dividing by 2, we get

x2 – 5x – 6 = 0

⇒ x2 – 6x + x – 6 = 0

⇒ x(x – 6) + 1(x – 6) = 0

⇒ (x + 1) (x – 6) = 0

So, x + 1 = 0 or x – 6 = 0

Thus, we get

x = -1 or x = 6

But, putting x = -1 the L.H.S become negative. And we know that the square root function always gives a positive value.

∴ x = 6 is the only solution.

2. One root of the quadratic equation 8x+ mx + 15 = 0 is ¾. Find the value of m. Also, find the other root of the equation.

Solution

Given equation, 8x+ mx + 15 = 0

One of the roots is ¾, and hence it satisfies the given equation

So,

8(3/4)+ m(3/4) + 15 = 0

⇒ 8(9/16) + m(3/4) + 15 = 0

⇒ 18/4 + 3m/4 + 15 = 0

Taking L.C.M, we have

(18 + 3m + 60)/4 = 0

⇒ 18 + 3m + 60 = 0

⇒ 3m = – 78

⇒ m = -26

Now, putting the value of m in the given equation, we get

8x+ (-26)x + 15 = 0

⇒ 8x– 26x + 15 = 0

⇒ 8x2 – 20x – 6x + 15 = 0

⇒ 4x(2x – 5) – 3(2x – 5) = 0

⇒ (4x – 3) (2x – 5) = 0

So, 4x – 3 = 0 or 2x – 5 = 0

∴ x = ¾ or x = 5/2

3. Show that one root of the quadratic equation x2 + (3 – 2a)x – 6a = 0 is -3. Hence, find its other root.

Solution

Given quadratic equation, x2 + (3 – 2a)x – 6a = 0

Now, putting x = -3 we have

(-3)2 + (3 – 2a)( -3) – 6a = 0

⇒ 9 – 9 + 6a – 6a = 0

⇒ 0 = 0

Since, x = -3 satisfies the given equation -3 is one of the root of the quadratic equation.

x2 + (3 – 2a)x – 6a = 0

⇒ x2 + 3x – 2ax – 6a = 0

⇒ x(x + 3) – 2a(x + 3) = 0

⇒ (x – 2a) (x + 3) = 0

So, x – 2a = 0 or x + 3 =0

x = 2a or x = -3

Hence, the other root is 2a.

4. If p – 15 = 0 and 2x2 + px + 25 = 0: find the values of x.

Solution

Given equations, p – 15 = 0 and 2x2 + px + 25 = 0

Thus, p = 15

Now, using p in the quadratic equation, we get

2x2 + (15)x + 25 = 0

⇒ 2x2 + 10x + 5x + 25 = 0

⇒ 2x(x + 5) + 5(x + 5) = 0

⇒ (2x + 5) (x + 5) = 0

So, 2x + 5 = 0 or x + 5 = 0

Hence, x = -5/2 or x = -5

5. Find the solution of the quadratic equation 2x2 – mx – 25n = 0; if m + 5 = 0 and n – 1 = 0.

Solution

Given,

m + 5 = 0 and n – 1 = 0

So, m = -5 and n = 1

Now, putting these values in the given quadratic equation 2x2 – mx – 25n = 0, we get

2x2 – (-5)x – 25(1) = 0

⇒ 2x2 + 5x – 25 = 0

⇒ 2x2 + 10x – 5x – 25 = 0

⇒ 2x(x + 5) -5(x + 5) = 0

⇒ (2x – 5) (x + 5) = 0

So, 2x – 5 = 0 or x + 5 = 0

Hence, x = 5/2 or x = -5

6. If m and n are roots of the equation: 1/x – 1/(x-2) = 3: where x ≠ 0 and x ≠ 2; find m x n.

Solution

Given equation, 1/x – 1/(x-2) = 3

⇒ (x – 2 – x)/ (x(x – 2)) = 3

⇒ -2 = 3(x2 – 2x)

⇒ 3x2 – 6x + 2 = 0

Solving by using quadratic formula, we get

And, since m and n are roots of the equation, we have

m = (√3 + 1)/ √3 n = (√3 – 1)/ √3

So, m×n = (√3 + 1)/ √3 × (√3 – 1)/ √3 = [(√3)2 – 1]/ (√3)2

Thus, m × n = 2/3